Answer to Question #173478 in Statistics and Probability for ANJU JAYACHANDRAN

Question

7(b) Let X X Xn

, , , 1 2 K be random sample of size n from a distribution with probability 

density function 





θ < < θ >

=

θ−

,0 else where.

0, ,1 0

( )0,;

1 X X

f X Obtain a maximum likeyhood 

estimator of θ.

Answer 1

Given probability distribution function is-

$f(x;θ)=θ^{−1}_2e^{−\dfrac{(x−θ_1)}{θ_2}}$

. The likelihood function is the density function regarded as a function of θ.

$L(θ|x) = f(x|θ), θ ∈ Θ. (1)$

The maximum likelihood estimator (MLE),

$\hat{θ}(x) = arg maxL(θ|x).$

for $x>θ_1,θ_2>0$ Likelihood:

$L(θ)=L(θ∣X_1,X_2,…,X_n)=∏_{i=1}^n\dfrac{e^{\dfrac{−(x_i−θ_1)}{θ_2}}}{θ_2}$

factored out $θ_2$ as well as turned to product into a summation to get:

$=\dfrac{1}{θ^n_2}e^{∑^n_{i=1}}\dfrac{−(x_i−θ_1)}{θ_2}$

$ℓ(θ)=ln(L(θ))=\dfrac{1}{θ^n_2}e^{∑^n_{i=1}}\dfrac{−(x_i−θ_1)}{θ_2}$

$=ln(θ^{−n}_2)+∑_{i=1}^n−\dfrac{(x_i−θ_1)}{θ_2}$

   $=−nln(θ_2)−\dfrac{∑^n_{i=1}x_i−nθ_1}{θ_2}$

    $=−nln(θ_2)−\dfrac{∑^n_{i=1}x_i}{θ_2}+\dfrac{nθ_1}{\theta_2}$